Let \(p \colon \tilde{X} \to X\) be a continuous map. An open set \(U \subseteq X\) is evenly covered by \(p\) if \(p^{-1}(U)\) is a disjoint union of open sets \(V_\alpha\) in \(\tilde{X}\text{,}\) each mapped homeomorphically onto \(U\) by \(p\text{.}\) The collection \(\{V_\alpha\}\) is called a partition into sheets.
A continuous map \(p \colon \tilde{X} \to X\) is a covering map, and \(\tilde{X}\) a covering space of \(X\text{,}\) if every point \(x \in X\) has an open neighbourhood that is evenly covered by \(p\text{.}\)
Show that if \(X\) is connected and \(\tilde{X}\) is non-empty, then a covering map is surjective and all fibres \(p^{-1}(x)\) have the same cardinality. If \(|p^{-1}(x)| = n\text{,}\) we call \(\tilde{X}\) an \(n\)-sheeted covering.
